Two identical capacitors have the same capacitance C . One of them...

Total energy of two identical capacitors before connection.

${E}_{1} = \dfrac{1}{2}C{V}_{1}^{2} + \dfrac{1}{2}C{V}_{2}^{2}$ .....(i)

When negative end of one capacitor is connected to negative end of other capacitor and positive to positive, then such type of connection is known as parallel connection.

Hence, equivalent capacitance after connection is given by,

${C}_{eq} = C + C$

${C}_{eq} = 2C$

Common potential, $V = \dfrac{Totalcharge}{Totalcapacitance}$

$= \dfrac{{Q}_{1} + {Q}_{2}}{C + C}$

$= \dfrac{C{V}_{1} + C{V}_{2}}{2C}$

$= \dfrac{{V}_{1} + {V}_{2}}{2}$

$∴$ Total energy after connections

${E}_{2} = \dfrac{1}{2} ⋅ {C}_{eq} ⋅ V^{2}$

$= \dfrac{1}{2} ⋅ 2C{\left(\dfrac{{V}_{1} + {V}_{2}}{2}\right)}^{2}$

$= \dfrac{C}{4}{\left({V}_{1} + {V}_{2}\right)}^{2}$

$∴$ Decrease in energy $= {E}_{1} - {E}_{2}$

$= \dfrac{1}{2}C{V}_{1}^{2} + \dfrac{1}{2}C{V}_{2}^{2} - \dfrac{C}{4}{\left({V}_{1} + {V}_{2}\right)}^{2}$

= \dfrac{C}{4} \left[2{V}_{1}^{2} + 2{V}_{2}^{2} - \left({V}_{1}^{2} + {V}_{2}^{2} +2{V}_{1}{V}_{2}\right)\right]<br/><br/> =\dfrac{C}{4}\left[{V}_{1}^{2} + {V}_{2}^{2} -2{V}_{1}{V}_{2}\right] = \dfrac{C}{4}{\left({V}_{1} - {V}_{2}\right)}^{2}